Question: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{\sqrt{9x^2+2}}{2x-9}=$
Solution: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x$, let's divide by $x$. In the numerator, let's divide by $-\sqrt{x^2}$, since for negative values, $x=-\sqrt{x^2}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{\sqrt{9x^2+2}}{2x-9} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{\sqrt{9x^2+2}}{-\sqrt{x^2}}}{\dfrac{2x-9}{x}} \gray{\text{Divide sides by }x=-\sqrt{x^2}} \\\\ &=\lim_{x\to-\infty}-\dfrac{\dfrac{\sqrt{9x^2+2}}{\sqrt{x^2}}}{\dfrac{2x-9}{x}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}-\dfrac{\sqrt{\dfrac{9\cancel{x^2}}{\cancel{x^2}}+\dfrac{2}{x^2}}}{\dfrac{2\cancel{x}}{\cancel{x}}-\dfrac{9}{x}} \\\\ &=\lim_{x\to-\infty}-\dfrac{\sqrt{9+\dfrac{2}{x^2}}}{2-\dfrac{9}{x}} \\\\ &=\lim_{x\to-\infty}-\dfrac{\sqrt{9+0}}{2-0} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=-\dfrac{\sqrt{9}}{2} \\\\ &=-\dfrac{3}{2} \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{\sqrt{9x^2+2}}{2x-9}=-\dfrac{3}{2}$.